A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is.

In SHM, V = ω√(A² - x²)
Initially, V = ω√(A² - (2A/3)²) = ω√(5A²/9) = (Aω√5)/3
After tripling the speed, V' = 3V = Aω√5
V' = ω'√(A'² - x²) = ω'√(A'² - (2A/3)²) = Aω√5
Since the position is the same, x = 2A/3
(Aω√5)/3 = ω√(A² - (2A/3)²)
ω = Aω√5 / 3√(5A²/9) = ω
Let A' be the new amplitude. Then V' = ω√(A'² - (2A/3)²)
3V = ω√(A'² - (2A/3)²)
3(Aω√5)/3 = ω√(A'² - 4A²/9)
Aω√5 = ω√(A'² - 4A²/9)
Squaring both sides:
5A² = A'² - 4A²/9
A'² = 5A² + 4A²/9 = (45A² + 4A²)/9 = 49A²/9
A' = 7A/3